Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a 'nice' proof to show that $\det(E^T) = \det(E)$ where $E$ is an elementary matrix?

Clearly it's true for the elementary matrix representing a row being multiplied by a constant, because then $E^T = E$ as it is diagonal.

I was thinking for the "row-addition" type, it's clearly true because if $E_1$ is a matrix representing row-addition then it is either an upper/lower triangular matrix and so $\det(E_1)$ is equal to the product of the diagonals. If $E_1$ is an upper/lower triangular matrix, then $E_1^T$ is a lower/upper triangular matrix and so $\det(E_1^T) = \det(E_1)$ as the diagonal entries remain the same when the matrix is transposed.

How about for the "row-switching" matrix where rows $i$ and $j$ have been swapped on the identity matrix? Can we use the linearity of the rows in a matrix somehow?

Thanks for any help!

share|cite|improve this question
It is not true that $\det{M} = \mathrm{tr}\,M$ for upper or lower triangular matrices. – josh Mar 20 '13 at 1:35
@josh Sorry, I've been massively stupid and thought the trace was the product of the diagonals NOT the sum. – Noble. Mar 20 '13 at 1:37
@josh I've edited the main post, sorry for the confusion! – Noble. Mar 20 '13 at 1:39
no big deal, glad you understand more now too. – josh Apr 2 '13 at 2:28
up vote 2 down vote accepted

You can use the fact that switching two rows or columns of a matrix changes the sign of the determinant. Switching two rows of $E$ makes it diagonal, then switch the corresponding columns and you have $E^T$

share|cite|improve this answer
Ah right, yes. Also, because you've switched rows/columns twice you've essentially multiplied the determinant by $(-1)^2$ and hence it's the same! Many thanks. – Noble. Mar 20 '13 at 1:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.